Exergy: Mechanical Nuclear Physics Measures Pressure, Viscosity and X-Ray Resonance in K-Shell in a Classical Way

2.1 Definitions

Nuclear reaction velocity coefficient.

We will use an equation analogous to concentration equation of Physical Chemistry C=C0ekt, where k=p02η,is velocity coefficient, p0is the initial pressure of our fluid, ηthe dynamic viscosity and C0the initial concentration of energetic fluid molecules.

It is evident that, in equilibrium state we can write μr=kt, however, the Navier–Stokes equation precisely measures the behavior of the fluids out of equilibrium, so that: μr≠kt.

Fortunately, there is a single solution for out-of-equilibrium fluids, using the fixed-point theorem for implicit functions, 11+ekt−μr=2μr,the proof is proved in Theorem 1.

Attenuation coefficient.

We will use the known attenuation formula of an incident flux I0,for which I=I0e−μr. Where, I0initial flux and μattenuation coefficient of energetic molecules that enter into interaction and/or resonance with the target molecules, transmitting or capturing the maximum amount of energy [5].

Dimensional analysis and fluid elements.

We will define the respective dimensional units of each one of variables and physical constants that appear in the solution of the Navier–Stokes 3D equation [12, 13, 14, 16].

Nuclear decay Nt=N0e−kt.Where k=1T1/2[1/s], the velocity coefficient, and T1/2ground state half-life.

Kinematic viscosity ν=ηρ0,m2s.

Dynamic viscosity η,[pa.s], where pa represents pascal pressure unit.

Initial Pressure of out of equilibrium. p0,[pa].

Fluid density ρ0,kgm3, where kgis kilogram and m3cubic meters.

Logistic probability function, Pxyzt=11+ekt−μr,it is a real number 0≤P≤1.

Equilibrium condition, r=kμt=p02ρ0νμt=uet,[m].

Fluid velocity in equilibrium, ue,m/s.Protons, neutrons and alpha particles are the elements of the fluid.

Fluid field velocity out of equilibrium, u=−2νμ1−P∇r.[m/s]. All nuclear decay is a process out of nuclear equilibrium.

Position, r=x2+y2+z21/2,[m].

Attenuation coefficient, μ,[1/m].

Growth coefficient, k=p02ρ0ν=p02η,[1/s].

Concentration C=C01−PP.

Theorem 1The velocity of the fluid is given byu=−2ν∇PP,wherePxyztis the logistic probability functionPxyxt=11+ekt−μr,andppressure such thatp=p0P,both defined onxyz∈R3t≥0.The functionPis the general solution of the Navier Stokes equations, which satisfies conditions (1) and (2).

Proof. To verify condition (2), ∇.u=0,we must calculate the gradients and laplacians of the radius. ∇r=xryrzr,and ∇2r=∇.∇r=y2+z2+x2+z2+x2+y2x2+y2+z23/2=2r.

Replacing the respective values for the terms: ∇2rand ∇r2in the Eq. (7).

Where the gradient modulus of ∇P=μP−P2∇r, has the form ∇P2=μ2P−P22∇r2=μ2P−P22.

Simplifying for 1−P≠0, we obtain the main result of this paper, which represents a fixed point of an implicit function ftrwhere ftr=P−2μr=0. In Nuclear Physics, r0<r<1.2A1/3.

Eq. (10) has a solution according to the fixed-point theorem of an implicit function, and it is a solution to the Navier Stokes stationary equations, which are summarized in: ∇2P=2μ∇21r=0.Furthermore, it is the typical solution of the Laplace equation for the pressure of the fluid ∇2p=p0∇2P=0. Kerson Huang (1987).

To this point, we need to verify that Eq. (10) is also a solution of requirement (1), ∂u∂t+u.∇u=ν∇2u−∇pρ0. We will do the equivalence u=∇θafter we replace in Eq. (1). Taking into account that θ=−2νlnP,and that ∇θis irrotational, ∇×∇θ=0, we have: u.∇u=∇θ.∇∇θ=12∇∇θ.∇θ−∇θ×∇×∇θ=12∇∇θ.∇θ,and ∇2u=∇∇.u−∇×∇×u=∇∇.∇θ−∇×∇×∇θ=∇∇2θ.Simplifying terms in order to replace these results in Eq. (1) we obtain

The explicit form of velocity is u=−2μν1−P∇r.Next, we need the partial derivative ∂u∂t

After replacing the last four results u.∇u,∇2u, ∂u∂tand −∇pρ0in Eq. (1) we obtain (11).

The Eq. (11) is equivalent to Eq. (1). After obtaining the term ∇P2P2from the incompressibility equation ∇∇2θ=−2ν∇−∇P2P2+∇2PP=0and replacing in Eq. (11).

Eq. (10) simultaneously fulfills requirements (1) expressed by Eq. (12) and requirement (2) expressed by Eq. (7), for a constant k=p02ρ0ν=p02η.Moreover, according to Eq. (10), the probability P=2μrwhich allows the Laplace equation to be satisfied: ∇2P=2μ∇21r=0.In other words, the Navier–Stokes 3D equation system is solved.▪

Implicit Function.

An implicit function defined as (10), ftr=11+ekt−μr−2μr=0has a fixed point trof R=tr0<a≤t≤b0<r<+∞, where mand Mare constants, such as: m≤M.Knowing that the partial derivative exists: ∂rftr=νP1−P+2μr2we can assume that: 0<m≤∂rftr≤M.If, in addition, for each continuous function φin abthe composite function gt=ftφtis continuous in ab,then there is one and only one function: r=φtcontinuous in ab, such that ftφt=0for all tin ab.

Theorem 2An implicit function defined as(10)ftr=11+ekt−μr−2μr=0has a fixed pointtrofR=tr0<a≤t≤b0<r<+∞. In this way, the requirements (1) and (2) are fulfilled.

Proof. Let Cbe the linear space of continuous functions in ab, and define an operator T:C→Cby the equation:

Then we prove that Tis a contraction operator, so it has a unique fixed point r=φtin C. Let us construct the following distance.

Using the mean value theorem for derivation, we have

Where ϕtis situated between φtand ψt. Therefore, the distance equation can be written as:

Using the hypothesis 0<m≤∂rftr≤Mwe arrive at the following result:

with which we can write the following inequality:

Where α=1−mM. Since 0<m≤M,we have 0≤α<1. The above inequality is valid for all tof ab. Where Tis a contraction operator and the proof is complete, since for every contraction operator T:C→Cthere exists one and only one continuous function φin C,such that Tφ=φ. Using Eq. (10), which represents the fundamental solution of the Navier–Stokes 3D equation, we verify Eq. (2), which represents the second of the six requirements of an acceptable solution.▪

Proposition 3Requirement (3). The initial velocity can be obtained from:uxyz0=−2ν∇PP,where each of the componentsux,uyanduzare infinitely derivable.

Proof. Taking the partial derivatives of ∂xnxr,∂ynxrand ∂znxr.

Recalling the derivatives of special functions (Legendre), it is verified that there exists the derivative C∞.

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Physically, this solution is valid for the initial velocity, indicated by Eq. (4), where the components of the initial velocity are infinitely differentiable, and make it possible to guarantee that the velocity of the fluid is zero when r→∞[6, 7, 8, 9].

Proposition 4Requirement (4). Using the initial velocity of a moving fluid given byuxyz0=u0xyz=−2νμ1−P0xryrzr, it is evident that

Proof. Using the initial velocity of a moving fluid given by u0xyz=−2νμ1−P0xryrzr, we can find each of the components: ∂xαux0,∂yαuy0and ∂zαuz0.

For the three components x,y,zthe results of the partial derivatives are as follows:

Replacing Eq. (17) with the explanatory form of the Legendre polynomials, for the following terms ∂xα−11rand ∂xα1r.

Also, knowing that for each α≥0, the maximum value of Pα1=1.We can write the following inequality

Grouping terms for ∂xαxr2,∂yαyr2and ∂zαzr2we have the next expressions.

The module of ∂xαu0is given by ∂xαu0=∂xαxr2+∂yαyr2+∂zαzr21/2.Simplifying and placing the terms of Eq. (20) we have

Taking into consideration that xr≤1,yr≤1,zr≤1the last term ∂xαu0can be easily written that.

It is verified that there exists Cα=10α!2such that if r→0,then ∂xαu0→0.Thus, we proved requirement (4).▪

According to Mathematics, and giving an integral physical structure to the study, we need to prove that there are the spatial and temporal derivatives of the velocity and pressure components, satisfying the requirement (5).

Proposition 5Requirement (5). The velocity can be obtained from:uxyzt=−2ν∇PPand each of the componentsux,uyanduzare infinitely derivable.

Proof. Taking partial derivatives for ∂xnxr2,∂ynxr2and ∂znxr2.

Recalling the derivatives of special functions, it is verified that the derivative C∞exists. These derivatives appear as a function of the Legendre polynomials Pn..

There are the spatial derivatives nand the time derivative which is similar to Eq. (25).▪

Proposition 6Requirement (5). The pressure is totally defined by the equivalencepxyzt=p0Pxyztand is infinitely differentiable in each of its components.

Proof. Taking partial derivatives for ∂xn1r,∂yn1rand ∂zn1r, recalling the derivatives of special functions of Eq. (16), it is shown that the derivative C∞. We only have to find the time derivatives: ∂tnp0P=p0∂tnP. Using Eq. (21) for P, we have.

It is always possible to find the derivative ∂tnPas a function of the previous derivative, since the resulting polynomial of each derivative n−1is of degree n.▪

Proposition 7Requirement (6). The energy must be limited in a defined volume and fundamentally it must converge at any time, such thatt≥0.

Proof. We will use the explicit form of velocity given in Eq. (21)uxyzt=2νμ1−P∇r,to obtain the vector module: u2=4ν2μ21−P2. Rewriting Eq. (21), and applying a change of variable in: dxdydx=4πr2dr.

Making another change of variable dP=μP1−Pdr. Using (10), replacing r2=2μP2we have

Where radius r→∞,when t≥0,we have limr→∞P=limr→∞11+expktexpμr=P∞=1.Moreover, physically if r→r0≈0then t→0we have limr→0P=limr→011+expktexpμr=P0=12.Here, a probability 12represents maximum entropy.

In this way the value of the constant Cis C=32πν2μ.Verifying the proposition (6) completely. In general, Eq. (10) can be written ftr+r0=11+ekt−μr+r0−2μμr+r0=0and in this way discontinuities are avoided when r→0, but this problem does not occur since in the atomic nucleus r0<r<1.2A1/3is satisfied.▪

Lemma 8The irrotational field represented by the logistic probability functionPxyztassociated with the velocityu=−2ν∇PP, can produce vortices, due to the stochastic behavior of the physical variablesp0,η,μ.These stochastic variations are in orders lower than the minimum experimental value.

Proof. The implicit function representing the solution of the Navier–Stokes 3D equation, 11+e−μr−kμt=2μrdepends on the values of initial pressure p0, viscosity νand attenuation coefficient μ. Due to Heisenberg uncertainty principle, these parameters have a variation when we measure and use them, as is the case of the estimate of ξ=r−ϕtkμ. The function ϕtkμ=kμt,expressly incorporates these results, when −∞<ξ<+∞. The physical and mathematical realities are mutually conditioned and allow for these surprising results. For a definite tthere exist infinities xyzthat hold the relationship r=x2+y2+z21/2. Moreover, for a definite rthere are infinities tthat respect the fixed-point theorem and create spherical trajectories. When the physical variables k,μvary, even at levels of 1/100 or 1/1000, they remain below the minimum variation of the experimental value. We could try to avoid the existence of trajectories on the spherical surface, for which we must assume that the fluid is at rest or it is stationary, which contradicts the Navier–Stokes 3D equation, where all fluid is in accelerated motion ∂u∂t≠0. In short, if there are trajectories in the sphere as long as it is probabilistically possible, this is reduced to showing that the expected value of the radius Err≥0exists and it is finite.

Derivation of Err≥0.

The logistic density function for ξwhen Eξ=0and Varξ=σ2is defined by.

hξ=μexp−ξ1+exp−ξ2,where 1μ=σ3/πis a scale parameter. Given that r=ϕtp0ημ+ξfunction for ris then fr=exp−r−ϕ•/ττ1+exp(−r−ϕ•/τ2,to facilitate the calculations we put ϕ•=ϕtkμ=kμt. By definition, the truncated density for rwhen r≥0is given by frr≥0=frPr≥0for r≥0.Given that the cumulative distribution function for ris given by Fr=11+expkt−μr,it follows that Pr≥0=1−F0=expϕ•)1+expϕ•)=11+exp−ϕ•). The derivation of Err≥0then proceeds as follows:

We replaced in Eq. (29)dP=μP1−Pdrand r2=2μP2of this manner we obtain

where we have used the fact that

where the last equality follows from an application of the L’Hopital’s rule Pr≥0=limt→∞expkt1+expkt=1.▪

3.1 The nuclear force and the Navier Stokes force are related

Firstly, we will use the concepts of Classic Mechanics and the formulation of the Yukawa potential, Φr=g4πrA−1e−μrto find the nuclear force exerted on each nucleon at interior of the atomic core FN=−∇Φr. Also, replace the terms of the potential e−μr=1−PPand 1r=μ2Pby the respective terms already obtained in Eq. (10).

The general form of the Eq. (32), is a function of xyzt.

Secondly, we will obtain the Navier Stokes force equation given by:

Theorem 9The Nuclear Force and Navier Stokes Force are proportional inside the atomic nucleusFN=CFNS.

Proof.Eq. (34) rigorously demonstrated by theorems and propositions 1 through 8, represent the acceleration of a particle within the atomic nucleus. According to Classical Mechanics the force of Navier Stokes applied to a particle of mass m, would have the form:

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Proof. The nuclear force on its part would be calculated as follows FN=−∇Φr.

Replacing the term ∇P=μP−P2∇rof Eq. (7), we obtain

It is possible to write nuclear force as a function of speed.

Finally, we can show that the nuclear force and force of Navier Stokes differ at most in a constant C. Equating (35) and (37), we find the value gas a function of the parameters nuclear viscosity ν, attenuation μand growth coefficient of the nuclear reaction k, nucleon mass mand C≠1.

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3.2 Cross section and Golden ratio σ1σ2are important elements of the femtoscope

According to NIST and GEANT4 [17], current tabulations of μρrely heavily on theoretical values for the total cross section per atom, σtot, which is related to μρby the following equation:

In (Eq. 39), u=1.6605402×10−24gris the atomic mass unit (1/12 of the mass of an atom of the nuclide 12C)4.

The attenuation coefficient, photon interaction cross sections and related quantities are functions of the photon energy. The total cross section can be written as the sum over contributions from the principal photon interactions

Where σpeis the atomic photo effect cross section, σcohand σincohare the coherent (Rayleigh) and the incoherent (Compton) scattering cross sections, respectively, σpairand σtripare the cross sections for electron-positron production in the fields of the nucleus and of the atomic electrons, respectively, and σph.n. is the photonuclear cross section3,4.

We use data of NIST and simulations with GEANT4 for elements Z=11to Z=92and photon energies 1.0721×10−3MeV to 1.16×10−1MeV, and have been calculated according to:

The attenuation coefficient μof a low energy electron beam 10,100eVwill essentially have the elastic and inelastic components. It despises Bremsstrahlung emission and Positron annihilation.

3.3 The principle of the femtoscope explains that low energy X-rays produce resonance in K-shell

A resonance region is created in a natural way at the K-shell between the nucleus and the electrons at S-level. The condition for the photons to enter in the resonance region is given by ra≥rn+λ. This resonance region gives us a new way to understand the photoelectric effect. There is experimental evidence of the existence of resonance at K-level due to photoelectric effect, represented by the resonance cross section provided by NIST and calculated with GEANT4 for each atom. In the present work we focus on the resonance effects but not on the origin of resonance region.

The resonance cross section is responsible for large and/or abnormal variations in the absorbed radiation I2−I1.

Theorem 10Resonance region. The resonance cross section is produced by interference between the atomic nucleus and the incoming X-rays inside the resonance region, where the boundaries are the surface of the atomic nucleus and K-shell.

The cross section of the atomic nucleus is given by:

The photon cross section at K-shell depends on the wave length and the shape of the atomic nucleus:

Subtracting the cross sections (13) and (14) we have:

The resonance is produced by interactions between the X-rays, the K-shell electrons and the atomic nucleus. The cross sections corresponding to the nucleus is weighted by probability pn and should have a simple dependence of an interference term. This last depends on the proton radius rp or the difference between the nucleus and proton radius (rn−rp) according to the following relation

We note that left hand side of Eqs. (46) and (47) should have a factor larger than one due to resonance. The unique factor that holds this requirement is σ1σ2bWhere a,bconstants.

After performing some simulations it is shown that the thermal a represents the dimensionless Rydberg constant a=R∞=1.0973731568539∗107.

We use σ1Z,σ2Zfor represent cross section in resonance, R∞is the generalized Rydberg constant for all elements of periodic table, and Zatomic number.

Last equation with a R2=0.9935was demonstrated and constructed using the elements of the solution of the Navier Stokes equations.

This equation was obtained with a R2=0.9939, and indicates that the ratio of the effective sections fully explain each element of the periodic table.

This equation obtained for a R2=0.9996, complements the system of equations that allow to know the simulation values as a function of Zfor σ1Z,σ2Zand E∗Z, where σ1Z<σ2Z.The Femtoscope equations further demonstrate that energy E∗=minEis minimum and Shannon entropy S∗=maxSis maximum in resonance, because in equilibrium σ1Zi=σ2Zi.

The radius of the neutron can be obtained using Eq. (49) in the following way.

3.4 Navier Stokes equation and cross section in nuclear physics

The speed needs to be defined as u=−2ν∇PP,where Pxyztis the logistic probability function Pxyxt=11+ekt−μr,r=x2+y2+z21/2defined in xyz∈R3t≥0This Pis the general solution of the Navier Stokes 3D equations, which satisfies the conditions (50) and (51), allowing to analyze the dynamics of an incompressible fluid.

Where, u∈R3an known velocity vector, ρ0constant density of fluid and pressure p=p0P∈R.

With speed and pressure dependent on rand t.We will write the condition of incompressibility as follows.

Theorem 11The velocity of the fluid given by:u=−2ν∇PP,wherePxyztis6the logistic probability functionPxyxt=11+ekt−μx2+y2+z21/2,defined inxyz∈R3t≥0is the general solution of the Navier Stokes equations, which satisfies conditions(50)and(51).

Proof. Firstly, we will make the equivalence u=∇θand replace it in Eq. (50). Taking into account that ∇θis irrotational, ∇×∇θ=0, we have.